If y=frac{cos x}{1+sin x},then(a) frac{dy}{dx | vedclass

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(C) Given $y = \frac{\cos x}{1+\sin x}$.Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:$\frac{dy}{dx} = \frac{

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Both $a$ and $c$ are correct

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(C) Given $y = \frac{\cos x}{1+\sin x}$.Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:$\frac{dy}{dx} = \frac{(-\sin x)(1+\sin x) - (\cos x)(\cos x)}{(1+\sin x)^2}$$\frac{dy}{dx} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2} = \frac{-(\sin x + \sin^2 x + \cos^2 x)}{(1+\sin x)^2}$Since $\sin^2 x + \cos^2 x = 1$,we have:$\frac{dy}{dx} = \frac{-(1+\sin x)}{(1+\sin x)^2} = \frac{-1}{1+\sin x}$. This matches option $(a)$.Now,simplify $\frac{-1}{1+\sin x}$ using trigonometric identities:$1+\sin x = 1+\cos\left(\frac{\pi}{2}-x\right) = 2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$.Therefore,$\frac{dy}{dx} = \frac{-1}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} = -\frac{1}{2}\sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$. This matches option $(c)$.Thus,both $(a)$ and $(c)$ are correct.

If $y=\frac{\cos x}{1+\sin x}$,then
$(a)$ $\frac{dy}{dx}=\frac{-1}{1+\sin x}$
$(b)$ $\frac{dy}{dx}=\frac{1}{1+\sin x}$
$(c)$ $\frac{dy}{dx}=-\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$
$(d)$ $\frac{dy}{dx}=\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$

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If $y=\frac{\cos x}{1+\sin x}$,then$(a)$ $\frac{dy}{dx}=\frac{-1}{1+\sin x}$$(b)$ $\frac{dy}{dx}=\frac{1}{1+\sin x}$$(c)$ $\frac{dy}{dx}=-\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$(d)$ $\frac{dy}{dx}=\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$